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Possibility to simplify $\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{\pi }{{\sin \pi a}}} $

Given that $0 < a < 1$ how to evaluate by the method of residues $$ \int_{-\infty}^\infty {e^{ax} \over 1 +e^x } \; dx $$

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In particular, see Argon's answer to that question. –  Pedro Tamaroff Sep 5 '12 at 19:30
    
@PeterTamaroff how did you change the sum into integral? –  hasExams Sep 6 '12 at 4:23
    
I explain it in detail in the question. –  Pedro Tamaroff Sep 6 '12 at 13:11
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marked as duplicate by Pedro Tamaroff, William, Did, Noah Snyder, J. M. Oct 5 '12 at 12:58

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2 Answers

up vote 4 down vote accepted

Substituting $u=e^x$, so that $dx = \dfrac{du}{u}$, the integral becomes

$$\int_0^{\infty} \dfrac{u^{a-1}}{1+u}du$$

How might you solve this? You've tagged the question as homework, so I'll leave it here for now, but if you're still stuck, post in the comments.

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does this method work when $0<a<1$ ? this needs a little editing and picture of contour is missing –  hasExams Sep 5 '12 at 16:18
    
@testuser: Yes. –  Clive Newstead Sep 5 '12 at 16:19
    
I must be an idiot :((( –  hasExams Sep 5 '12 at 16:20
    
@testuser: Possible, but unlikely! Just follow through that solution with $b=1$ all the way through. [Or work it out yourself $-$ it's not too hard using fairly standard complex methods.] –  Clive Newstead Sep 5 '12 at 16:22
    
yeah the answer is $ \pi \over \sin (a \pi )$ –  hasExams Sep 5 '12 at 16:24
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Recalling the $\beta$ function,

$$ \beta( n,m ) = \int_{0}^{1} u^{n-1}\,(1-u)^{m-1}\, du = \frac{\Gamma(n)\Gamma(m)}{\Gamma(m+n)} \,.$$

Using the substitution $ 1+u=\frac{1}{t} $ casts $ \int_0^{\infty} \dfrac{u^{a-1}}{1+u}du $ in terms of the $\beta$ function,

$$\int_0^{\infty} \frac{u^{a-1}}{1+u}du = \int _{0}^{1}\!{t}^{-a} \left( 1-t \right) ^{a-1}{dt} =\Gamma(a)\Gamma(1-a) = \frac{\pi}{\sin (a \pi )} \,. $$

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@Downvoter:I do not know what you are downvoting for. I gave a way to evaluate the integral in a way different from the complex techniques. –  Mhenni Benghorbal Sep 5 '12 at 21:04
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